Problem: Subtract. $\dfrac{8}{3} - \dfrac{1}{5} = $
Solution: Before we can subtract our fractions, they need to have the same denominator. $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\dfrac{8}{3}$ $\dfrac{1}{5}$ $\dfrac{8}{3}-\dfrac{1}{5}$ Let's look at the multiples of each denominator and see which multiples they have in common. Denominator Multiples ${3}$ $3, 6, 9, 12, \underline{15}, 18$ $5}$ $5, 10, \underline{15}, 20, 25$ The least common denominator is ${15}$. Let's use multiplication to make each fraction have a denominator of $15$. ${\dfrac{8}{3}}=\dfrac{{8} \times 5}{{3} \times 5} = {\dfrac{40}{15}}$ $\dfrac{1}{5}}=\dfrac{1} \times 3}{5} \times 3} = {\dfrac3}15}}$ Now, we can subtract ${\dfrac{40}{15}} - \dfrac{3}{15}}$. $\dfrac{40}{15}$ $\dfrac{3}{15}$ $\dfrac{40}{15} - \dfrac{3}{15}$ $=\dfrac{{40}-3}}{15}$ $= \dfrac{37}{15}$ ${\dfrac{8}{3}} - \dfrac{1}{5}} = \dfrac{37}{15}$ We can also write $\dfrac{37}{15}$ as $2\dfrac7{15}$.